485. Max Consecutive Ones

Given a binary array, find the maximum number of consecutive 1s in this array.

Example 1:

Input: [1,1,0,1,1,1]

Output: 3

Explanation: The first two digits or the last three digits are consecutive 1s.

    The maximum number of consecutive 1s is 3.

 

Note:

• The input array will only contain 0 and 1.
• The length of input array is a positive integer and will not exceed 10,000

 

 Solution 1: tranverse the array, use the counter cnt to sum the 1s: if the current number is 0, reset cnt=0; if the number is 1, cnt++. Then update the maximum number res.

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int max_cnt=0,cnt=0,size=nums.size();
        for (auto i:nums){
            if (i==1){
                max_cnt=max(++cnt,max_cnt);
            }
            else cnt=0;
        }
        return max_cnt;
    }
};

Solution 2: calculate sum(line 6)

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int res = 0, sum = 0;
        for (int num : nums) {
            sum = (sum + num) * num;
            res = max(res, sum);
        }
        return res;
    }
};